[Easy] Floor in a Sorted Array

Given a sorted array having no duplicates, arr[] and a value, x, find floor of x in given array.
Floor of x is the largest element in arr[] such that the element is smaller than or equal to x. If floor exists, then return index of it, else return -1.

Examples:

Input : arr[] = {1, 2, 8, 10, 11, 12, 19}, x = 5
Output : 1 
1 is index of 2. 2 is the largest element in arr[]
smaller than 5.

Input : arr[] = {1, 2, 8, 10, 11, 12, 19}, x = 20
Output : 6
19 is the largest element in arr[] smaller than 20.

Input : arr[] = {1, 2, 8, 10, 11, 12, 19}, x = 0
Output : -1
Since floor doesn't exist, output is -1.

Input:

The first line of input contains an integer T denoting the number of test cases.
The first line of each test case is N and x, N is the size of array. The second line of each test case contains N array elements.

Output:

Print index of floor of x if it exists, else print -1

Constraints:

1 ≤ T ≤ 500
1 ≤ N ≤ 1000
0 ≤ X ≤ 1000
1 ≤ arr[i] ≤ 10000

Example:

Input:
3
7 0
1 2 8 10 11 12 19
7 5
1 2 8 10 11 12 19
7 10
1 2 8 10 11 12 19

Output:
-1
1
3

def Floor(arr, x):
    for i in range(len(arr)):
        if arr[0] == x:
            return 0
        elif arr[0] > x:
            return -1
        elif arr[i] > x:
            return i-i
        elif arr[i] == x:
            return i
    return i

t = int(input())
for i in range(t):
    N = list(map(int, input().split()))
    x = N[1]
    arr = list(map(int,input().split()))
    print(Floor(arr, x))